Problem: Simplify the following expression: $y = \dfrac{2x^2+13x+21}{x + 3}$
Explanation: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(2)}{(21)} &=& 42 \\ {a} + {b} &=& &=& {13} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $42$ and add them together. The factors that add up to ${13}$ will be your ${a}$ and ${b}$ When ${a}$ is ${7}$ and ${b}$ is ${6}$ $ \begin{eqnarray} {ab} &=& ({7})({6}) &=& 42 \\ {a} + {b} &=& {7} + {6} &=& 13 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({2}x^2 +{7}x) + ({6}x +{21}) $ Factor out the common factors: $ x(2x + 7) + 3(2x + 7)$ Now factor out $(2x + 7)$ $ (2x + 7)(x + 3)$ The original expression can therefore be written: $ \dfrac{(2x + 7)(x + 3)}{x + 3}$ We are dividing by $x + 3$ , so $x + 3 \neq 0$ Therefore, $x \neq -3$ This leaves us with $2x + 7; x \neq -3$.